/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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YES

DP problem for innermost termination.
P =
  f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
  f5#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
  f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, 0, I14) 
  f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
R = 
  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
  f5(I5, I6, I7, I8, I9) -> f2(rnd1, rnd2, 0, I8, I9) [rnd1 = rnd2 /\ rnd2 = 0 /\ I9 <= 0 /\ 0 <= 100 - I8]
  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, 0, I14) 
  f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
  f1(I25, I26, I27, I28, I29) -> f2(I30, I31, 0, I28, I29) [I30 = I31 /\ I31 = 0 /\ 101 - I28 <= 0]

The dependency graph for this problem is:
  0 -> 2
  1 -> 4
  2 -> 1
  3 -> 4
  4 -> 3
Where:
  0) f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
  1) f5#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
  2) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, 0, I14) 
  3) f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
  4) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]

We have the following SCCs.
  { 3, 4 }

DP problem for innermost termination.
P =
  f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
R = 
  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
  f5(I5, I6, I7, I8, I9) -> f2(rnd1, rnd2, 0, I8, I9) [rnd1 = rnd2 /\ rnd2 = 0 /\ I9 <= 0 /\ 0 <= 100 - I8]
  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, 0, I14) 
  f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
  f1(I25, I26, I27, I28, I29) -> f2(I30, I31, 0, I28, I29) [I30 = I31 /\ I31 = 0 /\ 101 - I28 <= 0]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4,z5)] = 100 - z4 + -1 * 0
  NU[f3#(z1,z2,z3,z4,z5)] = 100 - z4 + -1 * 0

This gives the following inequalities:
    ==>  100 - I18 + -1 * 0 >= 100 - I18 + -1 * 0
  0 <= -1 + I24 /\ 0 <= 100 - I23  ==>  100 - I23 + -1 * 0 > 100 - (1 + I23) + -1 * 0  with  100 - I23 + -1 * 0 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 
R = 
  f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3]
  f5(I5, I6, I7, I8, I9) -> f2(rnd1, rnd2, 0, I8, I9) [rnd1 = rnd2 /\ rnd2 = 0 /\ I9 <= 0 /\ 0 <= 100 - I8]
  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, 0, I14) 
  f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 
  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23]
  f1(I25, I26, I27, I28, I29) -> f2(I30, I31, 0, I28, I29) [I30 = I31 /\ I31 = 0 /\ 101 - I28 <= 0]

The dependency graph for this problem is:
  3 -> 
Where:
  3) f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 

We have the following SCCs.